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How do you write #f(x) = 3x^2 + 6x + 1# in vertex form?

Answer 1

Isaiah Anthony

#f(x) = 3(x + 1)^2 - 2#

Complete the square:

#f(x) = 3(x^2 + 2x + m - m) + 1#

To find #m# we use #m = (b/2)^2 = (2/2)^2 = 1#

#f(x) = 3(x^2 + 2x + 1- 1) + 1#

#f(x) = 3(x^2 + 2x + 1) - 3 + 1#

#f(x) = 3(x + 1)^2 - 2#

Hopefully this helps!

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Answer 2

Nathan Axel

To write ( f(x) = 3x^2 + 6x + 1 ) in vertex form, first factor out the coefficient of ( x^2 ) from the ( x^2 ) and ( x ) terms, completing the square. This gives ( f(x) = 3(x^2 + 2x) + 1 ).

Now, to complete the square, add and subtract ((2/2)^2 = 1) inside the parentheses, but since you added 1 inside the parentheses, you need to subtract 3 as well outside to keep the expression equivalent. This gives:

( f(x) = 3(x^2 + 2x + 1 - 1) + 1 )

( f(x) = 3((x + 1)^2 - 1) + 1 )

( f(x) = 3(x + 1)^2 - 3 + 1 )

( f(x) = 3(x + 1)^2 - 2 )

Therefore, the vertex form of ( f(x) = 3x^2 + 6x + 1 ) is ( f(x) = 3(x + 1)^2 - 2 ).

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Answer from HIX Tutor

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